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I am studying for a qualifying exam, and have been working on this problem:

Let $D$ be a PID and let $P$ be a prime ideal of the polynomial ring $D[x]$. Suppose that $P$ contains a non-zero constant polynomial. Prove that $P$ can be generated by two elements.

I saw this, but I'm having trouble generalizing it for rings that aren't $\mathbb{Z}$. I know that $D[x]$ is also an integral domain (not a PID) so I don't get that $D[x]/P$ is a field, and I'm trying to understand the case where $P=(c,f(x),g(x))$ for some constant $c$ and two nonconstant polynomials $f(x),g(x)$. Does it work to replace $f$ and $g$ with their gcd? If so, can someone please explain why?

Thanks for your help.

user26857
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Erin
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If $P$ contains a non-zero element of $D$, then it contains a prime element of $D$ (why?), say $p$. Then $P/pD[x]$ is a prime ideal of $D[x]/pD[x]$. But $D[x]/pD[x]$ is isomorphic to $(D/pD)[x]$ (why?) and $D/pD$ is a field, so $D[x]/pD[x]$ is a PID. This shows that $P/pD[x]$ is principal. It follows that $P$ can be generated by two elements.

user26857
  • 52,094
  • @user26857 Thanks for your answer. I think I have most of the argument now.. Since $D$ is a UFD every nonzero element is a product of unit and irreducibles (primes) so $P$ must contain a prime element of $D$. And there exists a homomorphism from $D[x]$ to $(D/pD)[x]$ with kernel $pD[x]$ giving us the isomorphism. I'm just not sure how to get from $P/pD[x]$ principal to $P$ generated by two elements? If we let the generator of $P/pD[x]$ be $\alpha + pD[x]$ then is $P=(\alpha, c)$? – Erin Dec 30 '21 at 01:43
  • Yes. ${}{}{}{}$ – user26857 Dec 30 '21 at 04:35
  • Just for the record: if $P\cap D=(0)$, then $P$ is principal. – user26857 Dec 30 '21 at 04:44